Question

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1100 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 50 volts?
1.3e-8 is wrong

Answer 1

Capacitance of the deflection plate is given by,

C = 0*A/d

here, 0 = 8.85*10^-12

A = area = 10*2 cm^2 = 2*10^-3 m^2

d = gap distance = 1 mm = 1*10^-3 m

So, C = (8.85*10^-12)*(2*10^-3)/(1*10^-3)

C = 17.7*106-12 F = 17.7 pF

Now, Charge(Q) = C*V

given, V = potential difference = 100 volt

So, Q = (17.7*10^-12)*100

Q = 1.77*10^-9 C = 1.77 nC

Now, in RC circuit:

Potential difference across capacitor is given by,

Vc = V0*(1 - e^(-t/(RC)))

here, V0 = maximum voltage = 100 volt

Vc = 50 volts

R = 1100 ohm

So, 50 = 100*(1 - e^(-t/(1100*17.7*10^-12)))

100*e^(-t/(1100*17.7*10^-12)) = 100-50 = 50

e^(-t/(1100*17.7*10^-12)) = 0.50

t = -(1100*17.7*10^-12))*ln(0.5)

t = 1.35*10^-8 sec.

t = 13.5 ns

Answer WILL be 1.35*10^-8 sec, try your answer in nano sec unit and see if that works. Other ask your TA about this, as I'm 100% sure answer will be 1.35*10^-8 sec.